Question

A large steel wheel is to be fitted onto a shaft of the same material. At $27°C$ the outer diameter of the shaft is $8.70cm$ and the diameter of the central hole in the wheel is $8.69cm$. The shaft is cooled using dry ice. At what temperature of the shaft does the wheel slip on the shaft? Assume the coefficient of linear expansion of the steel to the constant over the required temperature range. ${\alpha }_{\mathrm{steel}}=1.20\times {10}^{-5}{K}^{-1}$.

Answer 1

Step-1: Given data

${\mathrm{T}}_1=27°\mathrm{C}$

Diameter of shaft ${\mathrm{l}}_1=8.70\mathrm{cm}$

Diameter of central hole ${\mathrm{l}}_2=8.69\mathrm{cm}$

The coefficient of linear expansion of the steel is ${\alpha }_{\mathrm{steel}}=1.20\times {10}^{-5}{K}^{-1}$.

Step 2: Determine the change in diameter when the wheel slips on the shaft

If the shaft does wheel slip on the shaft, then change in diameter, $∆l=l_2-l_1$.

$$\begin{gathered} \Rightarrow ∆l=\left(8.69-8.70\right)cm \\ \Rightarrow ∆l=-0.01cm \end{gathered}$$

Step 3: Determine the temperature at which the wheel slips on the shaft

Let the final temperature be $T_2$.

Thus, the change in diameter can be given by $∆l=l_1\alpha \left(T_2-T_1\right)$.

$$\begin{gathered} \Rightarrow -0.01=8.70\times 1.20\times {10}^{-5}\left(T_2-300\right) \\ \Rightarrow \frac{-0.01}{8.70\times 1.20\times {10}^{-5}}=\left(T_2-300\right) \\ \Rightarrow \left(T_2-300\right)=-95.79 \\ \Rightarrow T_2=300-95.79 \\ \Rightarrow T_2=204.21 \end{gathered}$$

Thus, the final temperature is $T_2=204.21K$.

$$\begin{gathered} \Rightarrow T_2=\left(204.21-273\right)°C \\ \Rightarrow T_2=-68.79°C \end{gathered}$$

Hence, the temperature of the shaft at which the wheel slip on the shaft is $-68.79°C$.